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Nezavi [6.7K]
3 years ago
7

Superman is riding his bicycle. At the base of a hill, he has a specific amount of kinetic energy. Then he coasts up the hill wi

thout pedaling. He comes to a stop at the top of the hill. Why is Superman’s potential energy at the top of the hill less than his kinetic energy at the bottom of the hill?
A. The hill did work on the bicycle as he went up the hill.
B. The rest of the energy is stored in the muscles of his body.
C. He still has kinetic energy while stopped on the top of the hill.
D. Friction and air resistance created heat on his trip up the hill.
Physics
1 answer:
algol133 years ago
7 0

Answer:

D. Friction and air resistance created heat on his trip up the hill.

Explanation:

Energy transformation from one form to another is not 100% efficient. This is the postulate of the first law of thermodynamics.

Most of the energy transformation is not purely 100%.

When energy is transformed, some are usually wasted.

  • In this case, in moving from bottom up, Superman produced some heat and encountered air resistance.
  • To reach the top, he must have overcome the resistance and produce enough heat to power him through.
  • This reduces the amount of potential energy that should have been the same as the kinetic energy down below.

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What is the SI unit for intensity?
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100mL of 4°C water is heated to 37 °C . Assume the density of the water is 1g/mL. The specific heat of water is 4.18 J/g(°C). Wh
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3 years ago
A high voltage transmission line with a resistance of 0.51 Ω/km carries a current of 1099 A. The line is at a potential of 1300
Illusion [34]
<h2>Answer:</h2>

\boxed{P=96.09MW}

<h2>Explanation:</h2>

First of all, we need to figure out what is the resistance in that line. In this problem, the total resistance is not given directly, but we can calculate it because we know it in terms of 0.51 Ω/km and since the distance from the power station to the city is 156km, then:

R_{line}=0.51 \frac{\Omega}{km}.156km \\ \\ R_{line}=79.56\Omega

So we can calculate the power loss as:

P=I^2R \\ \\ Where: \\ \\I=1099A \\ \\ P=(1099)^2(79.56) \\ \\ P=96092647.56W \\ \\ Remember \ that \ 1MW=10^6W \ So: \\ \\ P=96092647.56W(\frac{1M}{10^6}) \\ \\ \boxed{P=96.09MW}

Finally, the power loss due to resistance in the line is 96.09MW

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